Problem: What is the extraneous solution to these equations? $\dfrac{x^2}{x + 7} = \dfrac{x + 56}{x + 7}$
Answer: Multiply both sides by $x + 7$ $ \dfrac{x^2}{x + 7} (x + 7) = \dfrac{x + 56}{x + 7} (x + 7)$ $ x^2 = x + 56$ Subtract $x + 56$ from both sides: $ x^2 - (x + 56) = x + 56 - (x + 56)$ $ x^2 - x - 56 = 0$ Factor the expression: $ (x + 7)(x - 8) = 0$ Therefore $x = -7$ or $x = 8$ At $x = -7$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -7$, it is an extraneous solution.